Body movement question v(t), a(t), x(t)

  • Thread starter Thread starter mystmyst
  • Start date Start date
  • Tags Tags
    Body Movement
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
mystmyst
Messages
57
Reaction score
0
I don't learn in English so I may phrase words incorrectly. Just let me know if something doesn't make sense

Homework Statement



A body moves in one dimension where his speed is dependent on time:
[tex]v(t) = t^3-10t^2+16t, t \geq0[/tex]. v is in m/s amd t is in s.

a) what is it's initial speed?
b) what is it's initial acceleration?
c) find x(t) if it's known that x(0)=0
d) at what time does the body pass through the origin (x=0)?
e) find the extreme values (max and min) of the distance between the body and the origin, from the beginning of its movement and until the last time it passes through the origin

The Attempt at a Solution



a) initial speed is at t=0, so v=0 m/s
b) initial acceleration is at t=0 [tex]\rightarrow a(t) = 3t^2-20t+16 \rightarrow t=0, a=16 m/s^2[/tex]
c) [tex]x(t) = \frac{1}{4}t^4 - \frac{10}{3}t^3 + 8t^2[/tex] ------not sure about (c)
d) [tex]\frac{1}{4}t^4 - \frac{10}{3}t^3 + 8t^2=0 \rightarrow t=0s, t=10.194s, t=3.139s[/tex]
e) I am not sure about this one: wouldn't the min distance be 0m. since if x=0, there is no distance. and what do I do for the max distance? how do I know what the max distance is?

Thanks!
 
on Phys.org
You can find the distance extrema by looking for when velocity=0. Everything else looks good.